How to expand $\frac{1}{n^a(n+k)^b}$ using partial fraction decomposition?

Question

Is it possible to decompose $\displaystyle\frac{1}{n^a(n+k)^b}$ into finite summation?

where $a,b,n,k\in Z^{+}$ and $a+b$ is odd.

What I tried is converting the fraction to double integral:

$\displaystyle\frac{(-1)^{a+b}}{(a-1)!(b-1)!}\int_0^1\int_0^1 x^{n-1}y^{n+k-1}\ln^{a-1}(x)\ln^{b-1}(y)dxdy$ which I found useless.

Any method or reference is much appreciated.

Answer 1

In this paper, page 284, we have

$$\frac{1}{x^s(1-x)^t}=\sum_{j=1}^s\binom{s+t-j-1}{s-j}\frac1{x^j}+\sum_{j=1}^s\binom{s+t-j-1}{t-j}\frac1{(1-x)^j},$$

where $s,t \ge0, s+t\ge1.$

Replacing $x$ with $-x/n$ gives

$$\frac{(-n)^s}{x^s(1+x/n)^t}=\sum_{j=1}^s\binom{s+t-j-1}{s-j}\frac{(-n)^j}{x^j}+\sum_{j=1}^s\binom{s+t-j-1}{t-j}\frac1{(1+x/n)^j},$$

or

$$\frac{1}{x^s(n+x)^t}=\sum_{j=1}^s\binom{s+t-j-1}{s-j}\frac{(-1)^{s+j}}{n^{s+t-j}x^j}+\sum_{j=1}^s\binom{s+t-j-1}{t-j}\frac{(-1)^s}{n^{s+t-j}(n+x)^j}.$$

Answer 2

The partial fraction of : $$\dfrac{1}{X^a(X + k)^b}$$ is : $$\dfrac{1}{X^a(X + k)^b} = \sum_{p = 1}^a \dfrac{\alpha_p}{X^p} + \sum_{p = 1}^b \dfrac{\beta_p}{(X + k)^p}$$

Answer 3

Polynomials $P(x)=x^a$ and $Q(x)=(x+k)^b$ are coprime for $k \ne 0$, so by Bézout's identity there exist $A(x),B(x)$ with $\deg A \lt b, \deg B \lt a$ such that $A(x)P(x)+B(x)Q(x)=1$. Then:

$$ \frac{1}{x^a(x+k)^b}=\frac{A(x)\,x^a+B(x)\,(x+k)^b}{x^a(x+k)^b}=\frac{A(x)}{(x+k)^b}+\frac{B(x)}{x^a} $$

Expanding $A(x)$ in powers of $(x+k)$ as $A(x) = \sum_{i=0}^{b-1} a_i(x+k)^i$, and with $B(x)=\sum_{i=0}^{a-1} b_ix^i$:

$$ \frac{1}{x^a(x+k)^b}= \sum_{i=0}^{b-1} \frac{a_i}{(x+k)^{b-i}} + \sum_{i=0}^{a-1} \frac{b_i}{x^{a-i}} $$

Answer 4

Yes, it's possible: $\displaystyle\frac{\left(-1\right)^{a+b-1}}{n^a\left(n+k\right)^b}=\sum_{i+j=a+b}\binom{i-1}{a-1}\frac1{k^i\left(n+k\right)^j}+\sum_{i+j=a+b}\binom{j-1}{b-1}\frac{\left(-1\right)^i}{n^ik^j}$

In other words, $\displaystyle\frac{1}{n^a\left(n+k\right)^b}=-\sum_{i=a}^{a+b}\binom{i-1}{a-1}\frac{\left(-1\right)^{a+b}}{k^i\left(n+k\right)^{a+b-i}}-\sum_{j=b}^{a+b}\binom{j-1}{b-1}\frac{\left(-1\right)^{a+b-j}}{n^{a+b-j}k^j}$

Define $s=a+b\;$ and $\;\nu\in\{i,j\}.$ Using Cauchy product, we can rewrite this identity in terms of formal power series:

$\displaystyle n^b\left(n+k\right)^a\sum_{s=0}^\infty\frac{(-1)^{s-1}x^s}{\left(n\left(n+k\right)\right)^s}=\left(\sum_{\nu=0}^\infty\binom{\nu-1}{a-1}\frac{x^\nu}{k^\nu}\right)\left(\sum_{\nu=0}^\infty\frac{x^\nu}{\left(n+k\right)^\nu}\right)+\left(\sum_{\nu=0}^\infty\binom{\nu-1}{b-1}\frac{x^\nu}{k^\nu}\right)\left(\sum_{\nu=0}^\infty\frac{\left(-x\right)^\nu}{n^\nu}\right)$

This identity can be easily checked by some elementary manipulations with power series.